lecture 5 internal forces in beams in plane bending

V. DEMENKO MECHANICS OF MATERIALS 2020

12/2/2020 4:59:08 PMW:\+МЕХАНИКА МАТЕРИАЛОВ W\++НМКД АНГЛ\082 LECTURES 2020\05 Plane Bending Deformation. Diagrams of Internal Forces.doc

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LECTURE 5 Internal Forces in Beams in Plane Bending Deformation. Diagrams of Internal Forces

Introduction

A bar subjected to bending is called a beam.

Let us consider a model of bending when a plane of bending load coincides with

one of cross-sectional principal planes. This type of bending is sometimes called

simple bending. Actually, it is plane bending. The examples of plane bending

deformation are shown on Fig. 1.

Fig. 1

It is usual to distinguish pure bending (Fig. 2) and transverse bending (Fig. 3).

Pure bending is a type of loading producing the only bending moment ( yM or zM ) in

cross-section of a rod. In transverse bending, both shearing force ( zQ or yQ ) and

bending moment ( yM or zM ) occur in a rod section. The examples of pure plane

bending and transverse plane bending are shown on Figs. 2 and 3, respectively.



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Fig. 2

Fig. 3

1 Types of beam supports



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There are three main types of beam supports:

1. Movable hinged (pin) support;

2. Immobile hinged support;

3. Rigidly fixed (built-in) support.

The force factors that appear in supports are called support reactions.



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A. Illustration of movable hinged (pin) support

A movable hinged support (Fig. 4) permits free axial displacement of the beam

on rollers that’s why the only one support reaction appears into it. This reaction is

directed perpendicularly to support plane.

B. Illustration of immovable hinged (pin) support

At immobile hinged support, the reaction is directed to the horizontal axis at an

arbitrary angle ; this reaction can be resolved into two reaction components along the

horizontal and vertical axes (see Fig. 5).

Fig. 4



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The first two types of supports permit free rotation of the beam in support cross-

section, and therefore, no support moments appear in them (Figs. 4, 5). Only a rigidly

fixed (built-in) support, which does not permit the beam cross-section to rotate, can

create a reactive support moment (Fig. 6).

C. Illustration of rigidly fixed support

A rigidly fixed support gives three reactions: RM , vR , hR (see Fig. 6).

As a rule, the beam is in equilibrium, and therefore, three equilibrium equations are

used to describe this balance:

0 xF , 0 zF , 0M . (1)

Equations (1) are used to determine support reactions.

Fig. 5

Fig. 6



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2 The Sign Conventions in Internal Forces Calculation by Method of Sections

Fig. 7

If an external force applied to the left of the section m-m is directed upwards, the

shearing force at the section is considered to be positive 0mQ , otherwise negative

0mQ . It means, that the forces tend to produce clockwise rotation (Fig. 7) create

0mQ ; the forces tend to produce counter-clockwise rotation create 0mQ .

Fig. 8

The rule of signs for determining the bending moment can be formulated as follows: a

moment in cross-section n-n is considered to be positive if it affects the beam in the

way it becomes convex downwards, and vice versa (Fig. 5).

3 Shear Force and Bending Moment Diagrams for Cantilevers

Let us determine the internal force factors in an arbitrary section. To do this, the

beam is cut mentally through the section and the unknown internal force factors Qz and



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My are applied in the section to produce an equilibrium of any part of the beam under

consideration.

The equilibrium equations for the cut-off portion of the beam are as follows:

0 zF , 0 yM . (2)

In all cases, the shearing force for a prismatic rod is equal to the algebraical sum of all

external forces projections on the plane of section lying on one side of the section (left

or right). The bending moment at a section may be regarded as the sum of moments, in

relevance to the transverse axis in the section, of all forces applied to one side of this

section (left or right). In designing of bending moment and shearing force diagrams

with one tip fixed it is possible not to find the reactions in rigidly fixed support. Beams

of this kind are commonly called cantilevers.

Rule We will lay off the values of the bending moments from tensioned fibers of a

bent beam.

Fig. 9

Example 1 Internal forces

induced by concentrated external

moment

Given: M, l.

R.D.: zQ x , yM x .

Equations of internal force factors

are: I-I: 0 x l

0IzQ x ,

IyM x M .

The diagrams are design on Fig. 9.



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Fig. 10

Example 2 Internal forces

induced by concentrated external

force

Given: F, l.

R.D: zQ x , yM x .

Equations of internal force factors

are: I-I: 0 x l

IzQ x F ,

0

0Iy x x l

M x Fx Fl

.

The diagrams are design on Fig. 10.



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Example 3 Internal forces induced by uniformly distributed external load

Given: q, l, where q is the intensity

of uniformly distributed loading.

R.D.: zQ x , yM x .

The resultant ( qR ) of the uniformly

distributed load q is equal to the area of

the diagram q(x) and this resultant

passes through the “center of gravity” of

this diagram.

The shearing force diagram is

represented by a straight line. As may

be seen from expression (3) the shearing

force Qz in portion I-I increases from 0

to ql (0 x l ):

00 .

Iz q

x x l

Q x R

qx ql (3)

The maximum value of the shearing

force occurs at the built-in support.

The sum of the moments of external

forces lying on one (right) side of the

section is

2 2

0

02 2 2

Iy q x l

x

x qx qlM x R

, (4)

where qx is the resultant of uniformly distributed load q of x length. This resultant

passes through the midpoint of the segment x. Consequently, the arm of the force is

x/2, and the moment of this force is equal to 2

2 2 q

x xR q .

The bending moment diagram is represented by parabola. The maximum value of the

bending moment occurs at the built-in support. The graphs zQ x and yM x are

design on Fig. 11.

Fig. 11



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Example 4 Internal forces induced by linearly distributed external load

with increased intensity

Given: qm, l (qm is maximum intensity

of distributed load).

R.D.: zQ x , yM x .

Let’s consider the part of distributed

load with the maximum magnitude q(x)

applied to the right part of the beam:

Using similarity of the triangles:

m

m

q x q xxq x

q l l,

I-I: 0 x l

2

0

02 2

Iy q x l

x

qx qlQ x R

l

3

0

10

3 6

Iy q x l

x

qxM x R x

l

2

6

ql.

The graphs zQ x and yM x are

design on Fig. 12.

Fig. 12



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Example 5 Internal forces created

by linearly distributed external load with

decreased intensity

Given: qm, l.

R.D: zQ x , yM x – ?

To simplify the solution of this problem let

us apply two equal in magnitude but

oppositely directed linearly distributed

loads of the same intensity. One load is

directed downwards and the other is

directed upward. Thus, we have static zero

in any cross-section. This method alloys to

use two previous examples to solve the

problem:

I-I: 0 x l

2

0

0 ,2 2 2

Iy x l

x

qx ql qlQ x qx ql

l

2 3 2

0

02 6 3

Iy x l

x

qx qx qlM x

l.

The graphs zQ x and yM x are design

on Fig. 13.

Fig. 13

MINISTRY OF EDUCATION AND SCIENCE OF UKRAINE

National aerospace university “Kharkiv Aviation Institute”

Department of aircraft strength

Course Mechanics of materials and structures

HOME PROBLEM 4

Graphs of Shear Force and Bending Moment Distribution in Plane Bending (Cantilevers)

Name of student:

Group:

Advisor:

Data of submission:

Mark:

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Shear force in a prismatic rod is

equal to the algebraic sum of all

external forces projections on the

cross-section lying on one side of the

section (left or right).

The bending moment at a section

is equal to the sum of moments, in

relevance to the transverse axis in the

section, of all external forces applied

to one side of the section (left or

right).

Solution

1. Accepting the sign conventions in internal forces calculating.

(a) for shear force (b) for bending moment

Fig. 1

2. Writing the equations of shear force and bending moment functions in an arbitrary

sections of each potion (see Fig. 2).

Note. Method of sections will be applied from the right free tip of the cantilever and the

cross-sections will be established at x-distance relative to origin of each potion.

0m mzQ 0m m

zQ 0m myM 0m m

yM

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I – I (0 ) :x a

0 2

( ) 0 20Iz

x xQ x qx

kN,

0 2

( ) 20 02

Iy

x x

xM x M qx

kNm.

II – II (0 ) :x b

0 2

( ) 20 20IIz

x xQ x qa

kN,

0 2

( ) 0 402

IIy

x x

aM x M qa x

kNm.

III – III (0 ) :x c

0 2

( ) 10 10IIIz

x xQ x qa P qx

kN,

Note, that the change of shear force sign within the boundaries of this section predicts the

bending moment extreme value, since the derivative of bending moment is equal to shear

force:

( ( ))( ) .

IIIy III

z

d M xqa P qx Q x

dx

Therefore, zero shear force and also zero bending moment derivative represent the point of

bending moment extreme value.

To find it, let us determine the coordinate of zero shear force ex and substitute it into

bending moment equation.

1 30 10 2( ) 0 ( ) 1m

10

IIIz e e eQ x qa P qx x P qa

q

.

0 2

( ) 40 402 2

IIIy

x x

q xM x M qa a x Px qx

kNm.

max

2

( ) 352 2

III III ey e e ey

qxqM x M M qa a x Px

kNm.

3. Designing the graphs of shear forces and bending moments. Positive shear forces will

be drawn upwards and vice versa. Bending moment graph will be drawn on tensile fibers

according to the sign convention mentioned above (see Fig. 1 right). The graphs are shown

on Fig. 2.

4. Applying the shear force and bending moment values in the rigid support as

corresponding reactions according to accepted sign conventions (see Fig. 2).

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Fig. 2

lecture 5 internal forces in beams in plane bending

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